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We have a point charge $q>0$ in vacuum at a distance $a$ from the center $M$ of an isolated metal sphere that has a total charge $-q$ and radius $R$. ($a>R$)

A) Use the method of images to find a charge configuration that has the same boundary conditions. (hint: you need 3 charges)

For question A I know that my point charge $q$ should stay in the same place and that I need to place 2 other point charges inside the sphere. But I can't figure out what the charge and place of these two charges should be.

My book (Introduction to Electrodynamics, Griffiths) has an example where the sphere isn't charged but they seem to magically find the answer. They place a new charge $q'=-\displaystyle\frac{R}{a}q$ at a distance $b=\displaystyle\frac{R^2}{a}$ but they don't explain how this is a logical configuration. Can someone please explain this to me, as I would like to know how to place my point charges for different situations and not just one.

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    $\begingroup$ Inside the sphere, there is a charge $+q$. Therefore, on the surface (metal is a conductor) there must be a charge that will make the sphere's overall charge $-q$. Hence, on the surface the charge is $-2q$. As for the E-field, use Gauss's Law. $\endgroup$
    – Shahar
    May 19, 2014 at 12:01
  • $\begingroup$ Is the charge inside the metal sphere, or outside? In other words, is $a<R$? $\endgroup$
    – Floris
    May 19, 2014 at 12:27
  • $\begingroup$ the charge is outside the sphere, forgot to copy that part of the question $\endgroup$ May 19, 2014 at 13:42

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It should be enough to find the solutions to the following two simpler problems:

1) A point charge outside a grounded conducting sphere (so potential=0 on the sphere): how do you place an image charge inside the sphere to reproduce the field outside? This is the example from Griffiths you quote in your question. A natural way to get it is to start with the question "if I have two point charges $q_1$ and $q_2$ separated by a distance $d$, where does the resulting potential vanish?". Once you realize this is on a sphere, you can work backwards to find the relationships between positions and charges in terms of the parameters of the sphere.

2) An isolated charged conducting sphere, of some known charge (with no other point charges): how can you place an image charge inside this to reproduce the field outside? (This should be fairly obvious).

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  • $\begingroup$ why do we put an image charge INSIDE the sphere ? doesn't that contradict the fact that we aren't allowed to put image charges inside the domain of interest ? $\endgroup$
    – Mather
    Feb 25, 2019 at 10:22

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